How many even three-digit numbers have distinct digits and have no digit $5$? How many even three-digit numbers have distinct digits and have no digit $5$?
The answer my teacher gave was $252$, but I don't see how she got that. I thought it would be $6\times 8 \times 5=240$ because the $3\text{rd}$ digit must be even $(0,2,4,6,8)$, the $2\text{nd}$ digit can't be $5$ or the $3\text{rd}$ digit $(10-2=8$ options) and the first digit can't be $0$, $5$, the $2\text{nd}$ digit, or the $3\text{rd}$ digit ($10-4=6$ options). Any way I look at it, the last digit has to be even and there are $5$ options for even digits, so the final answer must end in a $5$ or a $0$, not a $2$. Please help!
 A: No. ending with non-zero digit = $4 .7.7=196$.
No. ending with zero digit = $1 .8.7=56$
P.S. The product above is (third digit options) x (first digit) x (second digit)
A: I figured out a different way to do it actually. I see that you have to account for when the third digit is zero and non-zero because it changes the criteria for the first digit, but I considered this for the second digit as well. What I did was split it into three cases:
Last digit is zero: $7 \times 8 \times 1=56$ (second digit cannot be $5$ or $0$ : the first digit cannot be $5$, the second digit or the third digit)
Last digit is nonzero and second digit is zero: $7 \times 1 \times 4=28$ (first digit cannot be $5$, the second digit or the third digit)
Second and last digits are nonzero: $6 \times 7 \times 4=168$ (the second digit cannot be $5, 0$, or the last digit: the first digit cannot be $5,0$, the second digit or the third digit)
Addition principle:$56+28+168=252$
It gives the right answer, but is this reasoning sound?
A: So, one is asking for the number of injections $f:\{1,2,3\} \to \{0,1,2,3,4,6,7,8,9\}$ where $f(1) \neq 0$ and $f(3) \in \{0,2,4,6,8\}$.
Ignoring the last two conditions, one gets $_{9}P_{3}=504$ possible numbers (possibly odd and/or having a leading zero).
If one requires the first digit to be nonzero, then one would then need to subtract the number of injections $\{2,3\} \to \{1,2,3,4,6,7,8,9\}$ to get $504-{_{8}P_{2}}=504-56=448$.
Odd numbers still have not been eliminated yet. We have to eliminate the cases where the last digit is $1, 3, 7,$ or $9$.
Suppose that the last two digits ($f(2)$ and $f(3)$) are $0$ and $1$ respectively. Then, one must have $f(1) \in \{2,3,4,6,7,8,9\}$, eliminating $7$ odd numbers to reduce the count to $441$.
Similar considerations apply when the middle digit is still $0$, but the last digit is now $3, 7,$ or $9$. This eliminates $3 \cdot 7=21$ more odd numbers, reducing the count to $420$.
Now, let's move on to the case where the middle digit is nonzero. Fixing a last digit $d \in \{1,3,7,9\}$, one would then need to subtract the number of injections $\{1,2\} \to \{1,2,3,4,6,7,8,9\} \setminus \{d\}$.
Suppose that $d=1$. Then, subtracting the number of injections $\{1,2\} \to \{2,3,4,6,7,8,9\}$ gives $420-{_{7}P_{2}}=420-42=378$. Forty-two odd numbers with the last digit equal to $1$ have now been eliminated.
Similar considerations apply when $d$ is $3,7,$ or $9$. This eliminates $3 \cdot 42=126$ more odd numbers, giving a final count of $252$.
